Euler-Lagrange equation

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Homework Statement



Let [tex]P[/tex] be a rectangle , [tex]f_{0} : \partial P \rightarrow R)[/tex] continuous and Lipschitz, [tex]C_{0} = \{ f \in C^{2}(P) : f=f_{0} \ on \ \partial P \}[/tex]. and finally [tex]S : C_{0} \rightarrow R[/tex] a functional:

[tex]S(f) = \int^b_a (\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy)\,dx + \int^d_c (\int^a_b (\frac{\partial f}{\partial y})^{2}\,dx)\,dy[/tex].

Write Euler-Lagrange equation for S.

Homework Equations





The Attempt at a Solution



I tried writing: [tex]S(f) = \int^b_a (\int^d_c (\frac{\partial f}{\partial x})^{2} + (\frac{\partial f}{\partial y})^{2}\,dy)\,dx [/tex], so the proper Lagrangian would be [tex]L(x) = \int^d_c (\frac{\partial f}{\partial x})^{2} + (\frac{\partial f}{\partial y})^{2}\,dy[/tex].

Then the Euler-Lagrange equation should be [tex]\frac{d}{dx}\frac{\partial L}{\partial f^{'}_{x}} = 0 \leftrightarrow \frac{d}{dx}\frac{\partial }{\partial f^{'}_{x}}\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy = 0[/tex], ([tex]L^{'} = \int^d_c (\frac{\partial f}{\partial x})^{2}\,dy[/tex]) and now since [tex]\frac{dL^{'}}{dx} = \frac{\partial L^{'}}{\partial f_{x}^{'}}\frac{\partial f_{x}^{'}}{\partial x}[/tex], we can rewrite that as [tex]\frac{d}{dx}\frac{\frac{d}{dx}\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy}{f_{xx}^{''}} = 0 \leftrightarrow \frac{d}{dx}\frac{\int^d_c \frac{\partial}{\partial x}(\frac{\partial f}{\partial x})^{2}\,dy}{f_{xx}^{''}} = 0 \leftrightarrow \frac{d}{dx}\frac{\int^d_c 2f_{x}^{'}f_{xx}^{''}\,dy}{f_{xx}^{''}} = 0[/tex]. but what then?
 
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Answers and Replies

  • #2
Dick
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You are making this too hard. Let's call [tex]\frac{\partial f}{\partial x}=f_x[/tex] and [tex]\frac{\partial f}{\partial y}=f_x[/tex]. Then the form of the Euler-Lagrange equations for two independent variables is [tex]\frac{\partial L}{\partial f}-\frac{\partial}{\partial x} \frac{\partial L}{\partial f_x}-\frac{\partial}{\partial y}\frac{\partial L}{\partial f_y}=0[/tex] where [tex]L=(f_x)^2+(f_y)^2[/tex].
 

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